# Can you help me with this simplifying problem

• Apr 15th 2012, 05:25 AM
Capnzims
Can you help me with this simplifying problem
I just want to confirm that the answer to this problem

54^j / 27^j3

54 power j power 1 / 27 j power 3

I come up with this answer 54^j-2 or should i write it 54^j-j3. Whats the proper format?
• Apr 15th 2012, 06:01 AM
Prove It
Re: Can you help me with this simplifying problem
Quote:

Originally Posted by Capnzims
I just want to confirm that the answer to this problem

54^j / 27^j3

54 power j power 1 / 27 j power 3

I come up with this answer 54^j-2 or should i write it 54^j-j3. Whats the proper format?

\displaystyle \displaystyle \begin{align*} \frac{54^j}{27j^3} &= \frac{\left(2 \cdot 27\right)^j}{27j^3} \\ &= \frac{2^j \cdot 27^j}{27j^3} \\ &= \frac{2^j \cdot 27^{j-1}}{j^3} \end{align*}
• Apr 15th 2012, 07:22 AM
Capnzims
Re: Can you help me with this simplifying problem
The book im using says that this answer would work too please tell me if this is correct
2/j^2 is that wrong?
• Apr 15th 2012, 08:06 AM
skeeter
Re: Can you help me with this simplifying problem
Quote:

Originally Posted by Capnzims
The book im using says that this answer would work too please tell me if this is correct
2/j^2 is that wrong?

is the original expression $\displaystyle \frac{54^j}{27^{j^3}}$ or $\displaystyle \frac{54j}{27j^3}$ ?
• Apr 15th 2012, 08:31 AM
Prove It
Re: Can you help me with this simplifying problem
Quote:

Originally Posted by skeeter
is the original expression $\displaystyle \frac{54^j}{27^{j^3}}$ or $\displaystyle \frac{54j}{27j^3}$ ?

Or even \displaystyle \displaystyle \begin{align*} \frac{54^j}{\left(27^j\right)^3} \end{align*}?
• Apr 15th 2012, 08:32 AM
Capnzims
Re: Can you help me with this simplifying problem
the j's are exponents raised to the power of the number beside them.
so j to the first power and j cubed
• Apr 15th 2012, 08:48 AM
skeeter
Re: Can you help me with this simplifying problem
Quote:

Originally Posted by Capnzims
the j's are exponents raised to the power of the number beside them.
so j to the first power and j cubed

then $\displaystyle \frac{2}{j^2}$ would not be correct.

however, since the "book" says it's a valid simplification, leads me to believe the original expression was meant to be $\displaystyle \frac{54j}{27j^3}$